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3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?

Physics

2 answers:

meriva3 years ago

6 0

Moon was rotation Earth or sun
it's own orbit . universe is not end

marysya [2.9K]3 years ago

3 0

Answer:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

Explanation:

"Assume the weight move up at constant speed. With no net acceleration, the force applied is -(weight). Since the weight at height R is -P(R/x)² (minusbecause it's directed downward) the applied lifting force is P(R/x)², and the work done moving from x to x+dx is dW = P(R/x)² dx. Intetgrate this:"

W(x) = ?PR²/x² dx = -PR²/x + C

The work done moving from x=R to x=2R is:

W(2R) - W(R) = -PR²(1/(2R) - 1/R) = PR/2

(b) The work done moving from 2R to 3R is:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

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Which item is heavier, 50 kg of cotton or 50 kg of iron?

Kipish [7]

Answer:

They weight the same

Explanation:

6 0

2 years ago

The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe

Crank

Answer:

The particle which completes the given equation is : $_{54}^{138}\textrm{Xe}$

Explanation:

The given reaction is of a fission reaction:

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_Z^A\textrm{X}+2_0^1\textrm{n}$

Total mass on the reactant side is equal to the total mass on the product side:

239 + 1 = 100 +A+ 2

A = 138

Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:

94 + 1(0) = 40 + Z + 2(0)

Z = 54

So atomic number 54 id of Xenon.

The particle which completes the given equation is :

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_{54}^{138}\textrm{Xe}+2_0^1\textrm{n}$

3 0

3 years ago

Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t

Andre45 [30]

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction $,\mu_k=0.11$

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

$F=\mu_k N$

Where N= Normal=mg

$F=0.11\times 634=69.74 N$

Hence, the force is needed to pull the sled across the snow at constant speed=69.74 N

7 0

2 years ago

At one point in the circuit, there is an LED and a resistor in parallel with one another. If you measure the voltage drop across

kifflom [539]

Answer:

C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.

Explanation:

When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.

If it were a series connection, then the current would have been the same, but the voltage drop were another story.

7 0

2 years ago

WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of

nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

$A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km$